Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{-6r - 12}{r - 6} \times \dfrac{r^2 - 14r + 48}{-6r - 12} $
First factor the quadratic. $q = \dfrac{-6r - 12}{r - 6} \times \dfrac{(r - 6)(r - 8)}{-6r - 12} $ Then factor out any other terms. $q = \dfrac{-6(r + 2)}{r - 6} \times \dfrac{(r - 6)(r - 8)}{-6(r + 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -6(r + 2) \times (r - 6)(r - 8) } { (r - 6) \times -6(r + 2) } $ $q = \dfrac{ -6(r + 2)(r - 6)(r - 8)}{ -6(r - 6)(r + 2)} $ Notice that $(r + 2)$ and $(r - 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -6\cancel{(r + 2)}(r - 6)(r - 8)}{ -6\cancel{(r - 6)}(r + 2)} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $q = \dfrac{ -6\cancel{(r + 2)}\cancel{(r - 6)}(r - 8)}{ -6\cancel{(r - 6)}\cancel{(r + 2)}} $ We are dividing by $r + 2$ , so $r + 2 \neq 0$ Therefore, $r \neq -2$ $q = \dfrac{-6(r - 8)}{-6} $ $q = r - 8 ; \space r \neq 6 ; \space r \neq -2 $